Math F665: Homework 7

Due: March 7, 2016

Solutions

1. Olver 4.2.3 (a-c) (Samaneh) 2. Olver 4.2.3 (a-c) Write down the solutions to the following initial-boundary value problems for the wave equation in the form of a Fourier series: a) utt = u xx , u(t, 0) = u(t, π) = 0, u(0, x) = 1, ut (0, x) = 0; b) utt = 2u xx , u(t, 0) = u(t, π) = 0, u(0, x) = 0, ut (0, x) = 1; c) utt = 3u xx , u(t, 0) = u(t, π) = 0, u(0, x) = sin3 x, ut (0, x) = 0. Solution, part a: By seperation of variables, we find that ν(x) must be satisfy d2 ν − λν = 0, dx2

ν(0) = ν(π) = 0.

Case 1: λ = −ω2 < 0 ν(x) = a sin(ωx) + b cos(ωx), ν(0) = b = 0, ν(π) = a sin(ωπ) = 0 =⇒ ωπ = kπ =⇒ ω = k. Thus ν(x) = a sin(kx),

k = 1, 2, ...

Consequently, the general solution to the boundary value problem is ∞ [ ] ∑ u(t, x) = bk cos(kt) sin(kx) + dk sin(kt) sin(kx) , k=1

where ∫ 2 π bk = u(0, x) sin(kx) dx, π 0 ∫ 2 π dk = ut (0, x) sin(kx) dx. k 0 Hence in this problem, ∫

π 2 cos(kx) kπ 0 0 2 2 = − (cos(kπ) − 1) = − ((−1)k − 1), kπ kπ

2 bk = π

π

sin(kx) dx = −

Math F665: Homework 7

Due: March 7, 2016

while dk = 0. Therefore, the solution is the Fourier series ∞ [ ∑

u(t, x) =

k=1

] 2 k − ((−1) − 1) cos(kt) sin(kx) . kπ

For λ = 0 and λ > 0, there is no solution. Solution, part b: By seperation of variables, we find that ν(x) must be satisfy d2 ν λ − ν = 0, dx2 2

ν(0) = ν(π) = 0.

Case 1: λ = −ω2 < 0 ω ω ν(x) = a sin( √ x) + b cos( √ x), 2 2 ν(0) = b = 0, √ ω ω ν(π) = a sin( √ π) = 0 =⇒ √ π = kπ =⇒ ω = 2k. 2 2 Thus ν(x) = a sin(kx),

k = 1, 2, ...

Consequently, the general solution to the boundary value problem is ∞ [ ] ∑ √ √ u(t, x) = bk cos( 2kt) sin(kx) + dk sin( 2kt) sin(kx) , k=1

where 2 bk = π

∫

π

u(0, x) sin(kx) dx, ∫ π 2 dk = √ ut (0, x) sin(kx) dx. 2k 0 0

Hence in this problem, ∫ π π 2 2 cos(kx) dk = √ sin(kx) dx = − √ 2 0 2k 0 2k 2 2 = − √ (cos(kπ) − 1) = − √ ((−1)k − 1), 2 2k 2k2 while bk = 0. Therefore, the solution is the Fourier series u(t, x) =

∞ [ ∑ k=1

] √ 2 − √ ((−1)k − 1) sin( 2kt) sin(kx) . 2k2 2

Solutions

Math F665: Homework 7

Due: March 7, 2016

Solutions

For λ = 0 and λ > 0, there is no solution. Solution, part c: By seperation of variables, we find that ν(x) must be satisfy d2 ν λ − ν = 0, dx2 3

ν(0) = ν(π) = 0.

Case 1: λ = −ω2 < 0 ω ω ν(x) = a sin( √ x) + b cos( √ x), 3 3 ν(0) = b = 0, √ ω ω ν(π) = a sin( √ π) = 0 =⇒ √ π = kπ =⇒ ω = 3k. 3 3 Thus ν(x) = a sin(kx),

k = 1, 2, ...

Consequently, the general solution to the boundary value problem is ∞ [ ] ∑ √ √ u(t, x) = bk cos( 3kt) sin(kx) + dk sin( 3kt) sin(kx) , k=1

where 2 bk = π

∫

π

u(0, x) sin(kx) dx, ∫ π 2 dk = √ ut (0, x) sin(kx) dx. 3k 0 0

Hence in this problem, bk = = b1 = b3 =

∫ 2 π 3 2 6 sin(πk) sin x sin(kx) dx = − 4 π 0 π k − 10k2 + 9 [ 1 sin((k − 3)x) 3 sin((k − 1)x) 3 sin((k + 1)x) sin((k + 3)x) ] − + − + , 4π∫ k−3 k−1 k+1 k+3 2 π 3 3 sin x sin(x) dx = , π 0 4 ∫ π 2 1 sin3 x sin(3x) dx = − , π 0 4

while dk = 0. Therefore, the solution is the Fourier series u(t, x) =

√ √ 3 1 cos( 3t) sin x − cos(3 3t) sin(3x) . 4 4 3

Math F665: Homework 7

Due: March 7, 2016

Solutions

For λ = 0 and λ > 0, there is no solution. 3. Olver 4.2.4 (Harrison) 4. Olver 4.2.9 (Solution by Maxwell) Problem delayed. 5. Olver 4.2.14 (Solution by Tyler) Sketch the solution of the wave equation utt = u xx and describe its behavior when the initial displacement is the box function { 1, 1 < x < 2, u(0, x) = 0, otherwise, while the initial velocity is 0 in each of the following scenarios: a) on the entire line −∞ < x < ∞; b) on the half-line 0 ≤ x < ∞, with homogeneous Dirichlet boundary condition at the end; c) on the half-line 0 ≤ x < ∞, with homogeneous Neumann boundary condition at the end; d) on the bounded interval 0 ≤ x ≤ 5 with homogeneous Dirichlet boundary conditions; and e) on the bounded interval 0 ≤ x ≤ 5 with homogeneous Neumann boundary conditions. Solution, part a: This IBVP satisfies the conditions under which we found d’Alembert’s solution, so denoting u(0, x) = f (x) we find u(t, x) =

(x − t) + f (x + t) . 2

(1)

This solution is a left- and right-going wave, each with width 1, height 12 , and speed 1. Sketches of this solution at various times are found in Figure 1. Solution, part b: The solution to this IBVP starts as a left- and right-going wave, each with width 1, height 12 , and speed 1. The right-going wave continues this path for all time. The leftgoing wave, however, is reflected at the origin and reappears upside down with the same profile but moving to the right with speed 1. This wave continues on this path for the rest of time. Sketches of this solution at various times are found in Figure 2.

4

Math F665: Homework 7

Due: March 7, 2016

Figure 1: Sketches of (1) at t = 0, 1, 3.

Figure 2: Sketches of 4.2.14b at t = 0, 1.3, 1.7, 3.

5

Solutions

Math F665: Homework 7

Due: March 7, 2016

Solutions

Figure 3: Sketches of 4.2.14c at t = 0, 1.7, 3.5. Solution, part c: As in part (b), the solution to this IBVP starts with a left- and right-going wave, each with width 1, height 21 , and speed 1. The right-going wave continuous this path for all time. The left-going wave, however, is reflected at the origin and the reflection remains upright. This wave then travels to the right with speed 1 for the rest of time. Sketches of this solution at various times are found in Figure 3. Solution, part d: Again the solution to the IBVP starts as a left- and right-going wave, each with width 1, height 21 , and speed 1. The left-going wave reflects at the origin and the right-going wave reflects at x = 5, with each reflection reappearing upside down with a reflected profile but moving in the opposite direction it was before the reflection. They recombine into a wave with width 1 and height -1 at time t = 5, and then continue with the same profile as they had before the interaction. At the origin and x = 5, the wave will reflect and appear right-side up. This pattern continues for all time. Sketches of this solution at various times are found in Figure 4. Solution, part e: Again the solution to the IBVP starts as a left- and right-going wave, each with width 1, height 12 , and speed 1. The left-going wave reflects at the origin and the right-going wave reflects at x = 5, with each reflection remaining upright with a reflected profile but moving in the opposite direction it was before the reflection. They recombine into a wave with width 1 and height 1 at time t = 5, and then continue with the same profile as they had before the interaction. At the origin and x = 5, the waves will reflect again. This pattern continues for all time. Sketches of this solution at various times are found in Figure 5. 6

Math F665: Homework 7

Due: March 7, 2016

Figure 4: Sketches of 4.2.14d at t = 0, 1.25, 1.75, 3.25, 3.75, 5.

Figure 5: Sketches of 4.2.14e at t = 0, 1.75, 3.75, 5.

7

Solutions

Math F665: Homework 7

Due: March 7, 2016

Solutions

6. Olver 4.2.24 (Solution by Caleb) a) Explain how to solve the Neumann initial-boundary value problem ∂u ∂u ∂2 u ∂2 u ∂u = 2, (t, 0) = 0 = (t, 1), u(0, x) = f (x), (0, x) = g(x), 2 ∂t ∂x ∂x ∂x ∂t on the interval 0 ≤ x ≤ 1. b) Let

  x − 14 , 14 ≤ x ≤ 12 ,    3 − x, 12 ≤ x ≤ 34 , f (x) =  4    0, otherwise,

and g(x) = 0. Sketch the graph of the solution at a few representative times, and discuss what is happening. Is the solution periodic in time? If so, what is the period? c) Do the same when f (x) = 0 and g(x) = x. Solution, part a: The solution to the same problem but considering Dirichlet boundary conditions is u(t, x) =

∞ ∑

(ak cos(kπt) +

k=1

bk sin(kπt)) sin(kπx) k

(2)

where ak , bk are the coefficients of the Fourier sine series of f and g respectively. The only thing that changes when we swap Dirichlet conditions for Neumann conditions is that we must consider the cosine Fourier series of f and g, and modify (2) accordingly. Therefore, the solution to this Neumann IBVP is ∑ bk a0 b0 + t+ (ak cos(kπt) + sin(kπt)) cos(kπx), u(t, x) = 2 2 k k=1 ∞

(3)

where ak are the Fourier cosine coefficients of f and bk are the Fourier cosine coefficients of g. Solution, part b: For the given initial conditions f and g, the solution in the form of (3) is ( ) ( ) ( ) π π 3π ∞ 1 ∑ 4 cos 2 k − 2 cos 4 k − 2 cos 4 k + cos(kπt) cos(kπx) u(t, x) = 16 k=1 π2 k 2 The solution is graphed in Figure 5.1 for various times t. At time t = 0, we have f (x). As t increases, the “triangle” splits into two pieces moving left and right, where its height as two separate triangles is 1/32. At time .5, the peak reaches both boundaries, and is then reflected back reaching the same profile as it moved to the boundary. At time 1, u returns to its initial profile f , and the process repeats. This solution has a period of 1. 8

Math F665: Homework 7

Due: March 7, 2016

Solutions

0.25 0.20 0.15

t =0,1 t = 18 , 78 t = 13 , 23 t = 25 , 35 t = 12

0.10 0.05 0.00 0.05 0.0

0.2

0.4

0.6

0.8

1.0

Figure 5.1 Solution, part c: For the given initial conditions f (x) = 0 and g(x) = x, ak = 0 for all k as the Fourier coefficients of 0 and the Fourier cosine coefficients of g(x) = x are b0 = 1 bk =

2((−1)k − 1) , π2 k2

for k = 1, 2, 3, .... Then the solution in the form of (3) is 1 ∑ 2((−1)k − 1) u(t, x) = t + sin(kπt) cos(kπx). 2 π3 k3 k=1 ∞

The solution is graphed in Figure 5.2 for various times t. At time t = 0, u ≡ 0, and for t ∈ (0, 1), the right side of the wave travels up, and this drags the left side up. By the time t = 1, the “right” side of the wave has spread to the left, and u is constant. Then the wave reflects off both boundaries, but there is more momentum on the left side. For t ∈ (1, 2), the profile is the reflected profile of u for t ∈ (0, 1), but translated up 1. The solution is u ≡ 1 at t = 2. Then the process will repeat. However, u is not periodic as it is strictly increasing with time. The boundary conditions do not specify values at the boundary (and the wave is reflected back evenly), and since there was a positive initial velocity, u will continue to grow.

9

Math F665: Homework 7

Due: March 7, 2016

Solutions

1.4 1.2 1.0 0.8 0.6

t =0 t = 13 t = 23 t =1 t = 43 t = 53 t =2 t = 73

0.4 0.2 0.0 0.0

0.2

0.4

0.6

Figure 5.2

10

0.8

1.0